Writing a vector parallel and normal to a unit vector ##\hat n##

[brotherbobby]

Homework Statement

A vector $\vec A$ can always be expressed as the sum of a vector parallel to an arbitrary unit vector $\hat n$ and perpendicular to it.

(1) Split the vector $\vec A$ into the two parts,

(2) Verify the direction of each part as being tangential and normal, respectively.

Relevant Equations

Using the dot product, a vector can be projected along a unit vector chosen arbitrarily : $A_n = \vec A\cdot\hat n.$

Drawing : Let me begin by drawing an image of the problem which I show to the right. During the drawing, it made sense that this separation is possible and unique.

Shown in the image alongside, we are given a vector $\boldsymbol{\vec A}$ and a unit vector ${\color{red}{\boldsymbol{\hat n}}}$. Effectively then, we are also given the unit vector perpendicular to it ${\color{red}{\boldsymbol{\hat n}_{\perp}}}$. The vector $\boldsymbol{\vec A}$ is decomposed into two parts, along and perpendicular to the given vector :

\begin{equation}
\boldsymbol{\vec A} = \vec{A}_n+\vec{A}_{\perp n}
\end{equation}

Question : What are the vectors $\vec{A}_n\;,\;\vec{A}_{\perp n}=?$ in equation $(1)$ above?

Attempt : I could guess (see Relevant Equations above) that the vector parallel to ${\color{red}{\boldsymbol{\hat n}}}$ is $\vec A_n=(\vec A\cdot \hat n)\hat n.{\Large\checkmark}$ This matches with the answer given.

I have no clue as to how to go about with the other vector $A_{\perp n}\; \hat{n}_{\perp}$. If I carried out a cross product $\vec A_{\perp n}=\hat n\times\boldsymbol{\vec A}$, it would go out of the plane. That would not be fine according to my first doubt (see below).

Doubt(s) : I am afraid I have several.

(1) Is this problem valid only for 2 dimensions? (That's what it seems to me. In three dimensions, there is a plane perpendicular to ${\color{red}{\boldsymbol{\hat n}}}$ implying an indefinite number of unit vectors on the plane and perpendicular to the given unit vector, making the decomposition of $\boldsymbol{\vec A}$ non unique.)

(2) Even in two dimensions, have I drawn the perpendicular unit vector ${\color{red}{\boldsymbol{\hat n}_{\perp}}}$ correctly, keeping uniqueness in mind? I show what I mean in the image to the right.

(3) What is $\vec A_{\perp n}$?

A hint or a clue would be of help.

 

[BvU]

Question : What are the vectors $\vec{A}_n\;,\;\vec{A}_{\perp n}=?$ in equation $(1)$ above?

Same as in your drawing: $\vec{A}_n$ is the component of $\vec {\bf A}$ along $\hat n$ and $\vec{A}_{\perp n}$ is $\vec {\bf A} - \vec{A}_n$

[edit] Note that $A_n$ is a number and $\vec A_n=A_n\,\hat n$ is a vector

Doubt(s) : (1) Is this problem valid only for 2 dimensions?

The decomposition is valid in any number of dimensions.
(I.e. the relevant equation works fine in any number of dimensions. Your "Effectively then, we are also given the unit vector perpendicular to it" is only correct in 2 dimensions -- [edit] except, as you noticed, there is a choice)

(2) Even in two dimensions, have I drawn the perpendicular unit vector
correctly, keeping uniqueness in mind. I show what I mean in the image to the right.

You are supposed to treat the unit vectors as given.
The decomposition process is unique.

$\$

 

[brotherbobby]

You are supposed to treat the unit vectors as given.

Unit vector, you mean. Only one unit vector is given.

The decomposition process is unique.

I agree with you. There are two unit vectors as I show here but the decomposition of the perpendicular component can only be along one unit vector (marked $\hat n_{\perp}$) or another.

The decomposition is valid in any number of dimensions.

In three dimensions, for instance, there will be an infinity of perpendicular unit vectors $\hat n_{\perp}$ along the plane normal to the given unit vector $\hat n$. Are you saying that the perpendicular component of $\mathbf A$, which I am calling $\vec A_{\perp n}$ , which will pick one of them only?

Same as in your drawing: $\vec{A}_n$ is the component of $\vec {\bf A}$ along $\hat n$ and $\vec{A}_{\perp n}$ is $\vec {\bf A} - \vec{A}_n$

So let me see. The vector perpendicular to $\hat n$ is $\boxed{\vec A_{\perp n} = \mathbf A - (\vec A\cdot \hat n)\;\hat n}$ .

I can't reduce the expression on the R.H.S further. However, I can tell you this is not the answer given in the text.

Spoiler: Answer

$\boxed{\mathbf A = (\mathbf A\cdot \hat n)\hat n + \hat n \times (\mathbf A\times \hat n)}$

I copy and paste a snipped image of the answer from the text to the right.

I hope I am not violating anything.

 

[kuruman]

Note that according to the BAC - CAB rule, $$\hat n\times(\mathbf A\times \hat n)=\mathbf A(\hat n \cdot \hat n)-\hat n(\hat n\cdot \mathbf A)=\mathbf A-(\mathbf A\cdot \hat n)\hat n.$$ Thus, $$\mathbf A=(\mathbf A\cdot \hat n)\hat n+\hat n\times(\mathbf A\times \hat n)$$ is an identity.

 

[brotherbobby]

Note that according to the BAC - CAB rule, $$\hat n\times(\mathbf A\times \hat n)=\mathbf A(\hat n \cdot \hat n)-\hat n(\hat n\cdot \mathbf A)=\mathbf A-(\mathbf A\cdot \hat n)\hat n.$$

Where did you obtain the first term from : $\hat n\times(\mathbf A\times \hat n)$?

 

[kuruman]

Where did you obtain the first term from : $\hat n\times(\mathbf A\times \hat n)$?

I applied the BAC - CAB rule with $\mathbf A \rightarrow \hat n$, $\mathbf B \rightarrow \mathbf A$ and $\mathbf C \rightarrow \hat n.$

 

[vela]

In three dimensions, for instance, there will be an infinity of perpendicular unit vectors n^⊥ along the plane normal to the given unit vector n^. Are you saying that the perpendicular component of A, which I am calling A→⊥n , which will pick one of them only?

Assuming $\hat n$ and $\vec A$ are not collinear, the two vectors define a plane. $\vec A_{\perp n}$ and $\hat n_\perp$ have to lie in this plane.

 

[brotherbobby]

I applied the BAC - CAB rule with $\mathbf A \rightarrow \hat n$, $\mathbf B \rightarrow \mathbf A$ and $\mathbf C \rightarrow \hat n.$

I know what you did. I am asking where did you get the vector $\hat n\times(\mathbf A\times \hat n)$ from?

 

[renormalize]

I am asking where did you get the vector $\hat n\times(\mathbf A\times \hat n)$ from?

Do you know and understand the general 3D vector identity:
$$\vec{a}\times\left(\vec{b}\times\vec{c}\right)\equiv\left(\vec{a}\cdot\vec{c}\right)\vec{b}-\left(\vec{a}\cdot\vec{b}\right)\vec{c}\quad\text{?}$$

 

[brotherbobby]

Do you know and understand the general 3D vector identity:
$$\vec{a}\times\left(\vec{b}\times\vec{c}\right)=\left(\vec{a}\cdot\vec{c}\right)\vec{b}-\left(\vec{a}\cdot\vec{b}\right)\vec{c}\quad\text{?}$$

Yes. I know what's going on.
But this is not the time to apply this rule. Because it involves what's in the answer which I am supposed to obtain.
Let's imagine that we don't know the answer and proceed from there.

 

[Orodruin]

Unit vector, you mean. Only one unit vector is given.

View attachment 362612I agree with you. There are two unit vectors as I show here but the decomposition of the perpendicular component can only be along one unit vector (marked $\hat n_{\perp}$) or another.

In three dimensions, for instance, there will be an infinity of perpendicular unit vectors $\hat n_{\perp}$ along the plane normal to the given unit vector $\hat n$. Are you saying that the perpendicular component of $\mathbf A$, which I am calling $\vec A_{\perp n}$ , which will pick one of them only?



So let me see. The vector perpendicular to $\hat n$ is $\boxed{\vec A_{\perp n} = \mathbf A - (\vec A\cdot \hat n)\;\hat n}$ .

I can't reduce the expression on the R.H.S further. However, I can tell you this is not the answer given in the text.

Spoiler: Answer

View attachment 362614 $\boxed{\mathbf A = (\mathbf A\cdot \hat n)\hat n + \hat n \times (\mathbf A\times \hat n)}$

I copy and paste a snipped image of the answer from the text to the right.

I hope I am not violating anything.

The answer given is a special case for three dimensions. The cross product is not defined in dimensions other than three.

The decomposition is however valid in any number of dimensions.

 

[vela]

I know what you did. I am asking where did you get the vector $\hat n\times(\mathbf A\times \hat n)$ from?

Use the right-hand rule to figure out what's going on.

 

[kuruman]

Where did you obtain the first term from : $\hat n\times(\mathbf A\times \hat n)$?

I looked at $\mathbf A-(\mathbf A\cdot \hat n)\hat n$ which is the perp. component and I thought to myself, "How can I make this elegantly compact? Hmm, this looks like the right hand side of the BAC - CAB rule with A and C the same and equal to a unit vector." It's a leap of thought that humans can do but not AI, at least not yet.

Another line of reasoning is suggested by @vela in post #12.

 

[brotherbobby]

Use the right-hand rule to figure out what's going on.

Ok you mean the BAC CAB rule in reverse. That will require some thinking.

To derive or imagine the double cross product on the RHS looking at the two vectors with their scalar multipliers on the LHS.

One of those terms is in the standard form, which helps. But the other is the vector itself : $\mathbf A$. One has to be able to "journey back" : $\mathbf A = (\hat n \cdot \hat n) \mathbf A$.

 

[kuruman]

Ok you mean the BAC CAB rule in reverse. That will require some thinking.

I can't speak for @vela, but the other line of reasoning I alluded to in post #13 is this.

You want to write $\mathbf A$ as the sum of two vectors: $~\mathbf A = \mathbf A_n+\mathbf A_{\perp}.$
You know that the parallel component of the first vector is $~ A_n=\mathbf A\cdot \hat n=|\mathbf A|\cos\!\varphi$ where $\varphi <\pi$ is the angle between $\mathbf A$ and $\hat n.$
Clearly, $\mathbf A_n=(\mathbf A_n\cdot \hat n)\hat n.$
You know that the magnitude of the perpendicular component must be $~|\mathbf A|_{\perp}=|\mathbf A|\sin\!\varphi=|\mathbf A\times \hat n|$. As a vector, this cross product has the same magnitude as $~\mathbf A_{\perp}$ but is perpendicular to the plane defined by $\mathbf A$ and $\hat n.$ You want to put it in this same plane without changing its magnitude. How can you do that?

Answer: You cross it with unit vector $\hat n$ and use the right hand rule to figure out whether it's $(\mathbf A\times \hat n)\times \hat n$ or $\hat n\times(\mathbf A\times \hat n)$ which has direction that is perpendicular to both $\mathbf A$ and $\hat n$ either "this way" or "the other way."

 

[brotherbobby]

I can't speak for @vela, but the other line of reasoning I alluded to in post #13 is this.

You want to write $\mathbf A$ as the sum of two vectors: $~\mathbf A = \mathbf A_n+\mathbf A_{\perp}.$
You know that the parallel component of the first vector is $~ A_n=\mathbf A\cdot \hat n=|\mathbf A|\cos\!\varphi$ where $\varphi <\pi$ is the angle between $\mathbf A$ and $\hat n.$
Clearly, $\mathbf A_n=(\mathbf A_n\cdot \hat n)\hat n.$
You know that the magnitude of the perpendicular component must be $~|\mathbf A|_{\perp}=|\mathbf A|\sin\!\varphi=|\mathbf A\times \hat n|$. As a vector, this cross product has the same magnitude as $~\mathbf A_{\perp}$ but is perpendicular to the plane defined by $\mathbf A$ and $\hat n.$ You want to put it in this same plane without changing its magnitude. How can you do that?

Answer: You cross it with unit vector $\hat n$ and use the right hand rule to figure out whether it's $(\mathbf A\times \hat n)\times \hat n$ or $\hat n\times(\mathbf A\times \hat n)$ which has direction that is perpendicular to both $\mathbf A$ and $\hat n$ either "this way" or "the other way."

Brilliant argument. Better than using the BAC-CAB rule which you referred to earlier. Still, there's an element of brilliance in that too. Let me use that as a second way out.

I don't mean to close the thread, but as the OP, I should solve the problem.

Problem statement :

Solution : (2) above is trivial if (1) is known.

(1) Let's write the vector $\mathbf{\vec A} = \vec A_n+\vec A_{\perp n}$. The symbols and image alongside make it evident what they mean.

Now, $\vec A_n=(\mathbf{\vec A}\cdot\hat n)\hat n$, using the idea of projection of a vector $\vec A$ along an arbitrary direction $\hat n$.

Thus $\mathbf{\vec A}=(\mathbf{\vec A}\cdot\hat n)\hat n+\vec A_{\perp n}\Rightarrow \vec A_{\perp n} = \mathbf{\vec A}-(\mathbf{\vec A}\cdot\hat n)\hat n.\qquad (1)$

The second term on the LHS above looks like the second term of the BAC-CAB rule : $\small{\vec{a}\times\left(\vec{b}\times\vec{c}\right)=\left(\vec{a}\cdot\vec{c}\right)\vec{b}-\left(\vec{a}\cdot\vec{b}\right)\vec{c}}$.

The first term on the RHS of $(1)$, which is just the vector $\mathbf{\vec A}$, can be rewritten : $\mathbf{\vec A} = (\hat n\cdot\hat n)\;\mathbf{\vec A}\quad \color{brown}{\Large !}\qquad (2)$

Both terms of $(1)$ now are the RHS of the BAC-CAB rule. Thus the required vector $\vec A_{\perp n}$ is the double cross product of the two unit (given) vectors $\hat n$ along with the (given) vector $\mathbf{\vec A}$ in the right order. This turns out to be the LHS of $(1)$ $=\hat n\times(\mathbf {\vec A}\times \hat n)\qquad (3)$

Rewriting $(1)$ using $(2)$ and $(3)$ above, we get the answer : $\boxed{\mathbf{\vec A}=(\mathbf{\vec A}\cdot\hat n)\hat n+\hat n\times(\mathbf A\times \hat n)}$

 

[vela]

Ok you mean the BAC CAB rule in reverse.

Not at all. I meant what @kuruman explained above.

 

[kuruman]

. . . but as the OP, I should solve the problem.

I thought you had already solved it. Sorry for spoiling your fun.

 

[haruspex]

Generalising somewhat, suppose in an n-dimensional vector space we wish to represent the vector $\vec v$ as a linear combination of basis vectors $\vec x_1…\vec x_n$.
If the coefficients are $\lambda_1…\lambda_n$ then $\vec v=\Sigma \lambda_i \vec x_i$. If we write the basis vectors as a matrix $X$ and the coefficients as vector $\vec\Lambda$ then $\vec v=X\vec\Lambda$.
Since $X$ is necessarily non-singular, $\vec\Lambda=X^{-1}\vec v$.

 

[Herman Trivilino]

Homework Statement: A vector $\vec A$ can always be expressed as the sum of a vector parallel to an arbitrary unit vector $\hat n$ and perpendicular to it.

(1) Split the vector $\vec A$ into the two parts,

(2) Verify the direction of each part as being tangential and normal, respectively.
Relevant Equations: Using the dot product, a vector can be projected along a unit vector chosen arbitrarily : $A_n = \vec A\cdot\hat n.$

I have no clue as to how to go about with the other vector A⊥nn^⊥. If I carried out a cross product A→⊥n=n^×A→, it would go out of the plane. That would not be fine according to my first doubt (see below).

Why, suddenly, are you concerned with the cross product? Prior to that you mention only the dot product. And it's true that the dot product between two perpendicular vectors is zero. Isn't that all you need to know and understand here?

 

[renormalize]

Why, suddenly, are you concerned with the cross product?

Because he's trying to understand the answer as given in the book he's working from:

I can't reduce the expression on the R.H.S further. However, I can tell you this is not the answer given in the text.
$\boxed{\mathbf A = (\mathbf A\cdot \hat n)\hat n + \hat n \times (\mathbf A\times \hat n)}$

I copy and paste a snipped image of the answer from the text to the right.

 

[Orodruin]

I looked at $\mathbf A-(\mathbf A\cdot \hat n)\hat n$ which is the perp. component and I thought to myself, "How can I make this elegantly compact? Hmm, this looks like the right hand side of the BAC - CAB rule with A and C the same and equal to a unit vector." It's a leap of thought that humans can do but not AI, at least not yet.

Another line of reasoning is suggested by @vela in post #12.

I actually think $\vec A - (\vec A \cdot \hat n)\hat n$ is nicer than the double cross product. It also generalises to any dimension.

 

[kuruman]

I actually think $\vec A - (\vec A \cdot \hat n)\hat n$ is nicer than the double cross product. It also generalises to any dimension.

Perhaps. I think that, if one is given $\mathbf A$ and $\hat n$ and the goal is to find vector components $\mathbf A_{\parallel}$ and $\mathbf A_{\perp}$ such that $\mathbf A =\mathbf A_{\parallel}+\mathbf A_{\perp}$, one would have to find independent-looking expressions for the parallel and perpendicular vector components. If one follows your suggestion, one would end up with $$\mathbf A= (\mathbf A\cdot \hat n)\hat n +\mathbf A-(\mathbf A\cdot \hat n)\hat n=\mathbf A$$ which is not reassuring, I think.

 

[Orodruin]

Perhaps. I think that, if one is given $\mathbf A$ and $\hat n$ and the goal is to find vector components $\mathbf A_{\parallel}$ and $\mathbf A_{\perp}$ such that $\mathbf A =\mathbf A_{\parallel}+\mathbf A_{\perp}$, one would have to find independent-looking expressions for the parallel and perpendicular vector components. If one follows your suggestion, one would end up with $$\mathbf A= (\mathbf A\cdot \hat n)\hat n +\mathbf A-(\mathbf A\cdot \hat n)\hat n=\mathbf A$$ which is not reassuring, I think.

On the contrary. It is very reassuring that the components of $\vec A$ add up to $\vec A$. You should worry if they didn’t!

The more relevant conputation is showing that the orthogonal component is indeed orthogonal to $\hat n$ and that the parallel component has the same inner product with $\hat n$ as $\vec A$ itself.

 

[kuruman]

On the contrary. It is very reassuring that the components of $\vec A$ add up to $\vec A$. You should worry if they didn’t!

True, if I see = + - , I don't worry whether things add up.

It doesn't say anything interesting either.

 

[robphy]

I actually think $\vec A - (\vec A \cdot \hat n)\hat n$ is nicer than the double cross product. It also generalises to any dimension.

When learning relativity, this decomposition was one of my "a-ha" moments
because I could see the use of the vectors $\vec A$ and (not necessarily unit) $\vec n$ and the metric.
(see p16-17 of General Relativity (1972) at https://home.uchicago.edu/~geroch/Course Notes )
The cross-product form is secondary and is possibly useful in (3+1)-cases...
but is not helpful in (1+1)-cases.

In addition, this decomposition is used repeatedly in the Gram-Schmidt Process in linear algebra.
The cross-product is generally not useful here.

 

[Orodruin]

There is actually a generalization of the cross product to arbitrary dimensions that will do the same trick asthe cross product. However, it maps two vectors anti-symmetrically to an antisymmetric N-2 tensor. The end result can be expressed using the generalized Kronecker delta.